WOW! Somebody (maybe the owner) increased the unsolved puzzles prizes again by x10 😱
Now the puzzle #66 prize is 6.6 BTC, #67 is 6.7 BTC... and so on .... puzzle # 160 prize is 16 BTC now
👍🏼🥳
Topic: == Bitcoin challenge transaction: ~1000 BTC total bounty to solvers! ==UPDATED== - page 22. (Read 47899 times)
April 16, 2023, 02:31:48 AM
April 08, 2023, 02:31:24 PM
13zb1hQbw fhhEryPWBrAioLY5tiFLr hs5o - 0x319AFC9E3FF391B01
take a good look at the end of the address it is in no way identical to the 66 which is h5so no hs5o
13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
it is only the first 7 characters which are identical but several utilities can allow you to find some in this range with even more characters just for example
13zb1hQbWLiYzxkb5Qeiwr9ZZRHycheQBV
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZo5eQmmER3KXshr85jU
2E120C91F20C0B511
20d45a6a762458afee967c590479283af57e7c40
take a good look at the end of the address it is in no way identical to the 66 which is h5so no hs5o
13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5soit is only the first 7 characters which are identical but several utilities can allow you to find some in this range with even more characters just for example
13zb1hQbWLiYzxkb5Qeiwr9ZZRHycheQBV
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZo5eQmmER3KXshr85jU
2E120C91F20C0B511
20d45a6a762458afee967c590479283af57e7c40
i know that, and thats what im talking about men
, its located in the range of 200.. to 3fff.. 
if you find same 7 or 8 digits of the rmd160 that means its in that range bro, so, when i post the public key substracted from 125 to 110. i told you this is in the 110 bits because it start with the 4 same rmd160 of other wallet in 110, that means its in the same range of 110, like you said bro in 66 20d45a6a same start of wallet 66, and the range is 319afc.... so, do you understand what im talking ?Think of it this way; there are only 2^16 (or 16^4) combinations of the same 4 leading rmd160 characters; in all of the 2^160 combinations.
April 08, 2023, 02:17:13 PM
i know that, and thats what im talking about men , its located in the range of 200.. to 3fff.. if you find same 7 or 8 digits of the rmd160 that means its in that range bro, so, when i post the public key substracted from 125 to 110. i told you this is in the 110 bits because it start with the 4 same rmd160 of other wallet in 110, that means its in the same range of 110, like you said bro in 66 20d45a6a same start of wallet 66, and the range is 319afc.... so, do you understand what im talking ?
, its located in the range of 200.. to 3fff.. if you find same 7 or 8 digits of the rmd160 that means its in that range bro, so, when i post the public key substracted from 125 to 110. i told you this is in the 110 bits because it start with the 4 same rmd160 of other wallet in 110, that means its in the same range of 110, like you said bro in 66 20d45a6a same start of wallet 66, and the range is 319afc.... so, do you understand what im talking ?I don't understand anything, the start string of the address don't have any relationship with the range
April 08, 2023, 11:41:00 AM
13zb1hQbw fhhEryPWBrAioLY5tiFLr hs5o - 0x319AFC9E3FF391B01
take a good look at the end of the address it is in no way identical to the 66 which is h5so no hs5o 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
it is only the first 7 characters which are identical but several utilities can allow you to find some in this range with even more characters just for example
13zb1hQbWLiYzxkb5Qeiwr9ZZRHycheQBV
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZo5eQmmER3KXshr85jU
2E120C91F20C0B511
20d45a6a762458afee967c590479283af57e7c40
take a good look at the end of the address it is in no way identical to the 66 which is h5so no hs5o
13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5soit is only the first 7 characters which are identical but several utilities can allow you to find some in this range with even more characters just for example
13zb1hQbWLiYzxkb5Qeiwr9ZZRHycheQBV
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZo5eQmmER3KXshr85jU
2E120C91F20C0B511
20d45a6a762458afee967c590479283af57e7c40
i know that, and thats what im talking about men
, its located in the range of 200.. to 3fff.. if you find same 7 or 8 digits of the rmd160 that means its in that range bro, so, when i post the public key substracted from 125 to 110. i told you this is in the 110 bits because it start with the 4 same rmd160 of other wallet in 110, that means its in the same range of 110, like you said bro in 66 20d45a6a same start of wallet 66, and the range is 319afc.... so, do you understand what im talking ? April 08, 2023, 07:38:23 AM
13zb1hQbw fhhEryPWBrAioLY5tiFLr hs5o - 0x319AFC9E3FF391B01
take a good look at the end of the address it is in no way identical to the 66 which is h5so no hs5o 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so
it is only the first 7 characters which are identical but several utilities can allow you to find some in this range with even more characters just for example
13zb1hQbWLiYzxkb5Qeiwr9ZZRHycheQBV
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZo5eQmmER3KXshr85jU
2E120C91F20C0B511
20d45a6a762458afee967c590479283af57e7c40
take a good look at the end of the address it is in no way identical to the 66 which is h5so no hs5o
13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5soit is only the first 7 characters which are identical but several utilities can allow you to find some in this range with even more characters just for example
13zb1hQbWLiYzxkb5Qeiwr9ZZRHycheQBV
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3qZo5eQmmER3KXshr85jU
2E120C91F20C0B511
20d45a6a762458afee967c590479283af57e7c40
April 08, 2023, 01:21:39 AM
I can do 1 quintillion/sec 1,000,000,000,000,000,000 but the range of 110 alone is 649037107316853453566312041152512 or
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
with wich programm?
April 05, 2023, 06:40:38 PM
I can do 1 quintillion/sec 1,000,000,000,000,000,000 but the range of 110 alone is 649037107316853453566312041152512 or
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
i have the same speed 7terrak per sec thats 7 quantillion per sec , thats why i told you my speed is slow im running bsgs on 8 core cpu, so i cant complete the entire range in 1 week or month, thats why i ask for help here.
... i found a substracted key his hash160 start with the same 4 lettres as a hash160 of an adress in the range 110 bit, so, am i close to find it ?
...
...
No you aren't any closer
i understand, thanks for your help, i was thinking like that because in this post
https://bitcointalksearch.org/user/dextronomous-306088
Puzzle 66... ( 13zb1hQbWVsc2S7ZTZnP2G4undNNpdh5so )
That's a mockery ....!
13zb1hQbwfhhEryPWBrAioLY5tiFLrhs5o - 0x319AFC9E3FF391B01
That's a mockery ....!
13zb1hQbwfhhEryPWBrAioLY5tiFLrhs5o - 0x319AFC9E3FF391B01
my guess. it should be above this one,
00000000000000000000000000000000000000000000000319AFC9E3FF391B01
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q > ZsnspiitMKGTKUxzXus
KwDiBf89QgGbjEhKnhXJuH7LrciVrZi3q > (a) something
ripemd -- 20D45A6A76AEB187FB6E669463E929D7072AB330
13zb1hQbwfhhEryPWBrAioLY5tiFLrhs5o
they find same firsts caracteres hash160 of the puzzle 66 so logic its on that bit range 66
thats why when i found the same 4 hex of 110 bits of other 110 bit wallet i say that, im just asking the question and if in 2020 they cracked the 110 bit, if you have a good speed and you can search for the entire range in 1 hour you can try the public key, maybe im wrong but if its correct, you will earn smthing,
you can substract the bit 125 but you wont find a hash same as you have in a specific bit range start with same digits, its luck i think when i found it but maybe im not wrong ,, if we found hashes starts with same 4 digits in all bit ranges then im wrong, that information i dont have, i dont know how the hash160 works so, i just want you to help guys maybe there is smthing.I can do 1 quintillion/sec 1,000,000,000,000,000,000 but the range of 110 alone is 649037107316853453566312041152512 or
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
I have. I have over 33GB of tame points (DP 25) in the 110 bit range lol. But I wouldn’t chase down a 4 character h160 that’s for sure lol. six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
Also, the “1 hour” entire range scan had me laughing 😂
i dont know 1 hour or 1 day, but if you want to have a test of that public key, thanks
April 05, 2023, 05:54:03 PM
... i found a substracted key his hash160 start with the same 4 lettres as a hash160 of an adress in the range 110 bit, so, am i close to find it ?
...
...
No you aren't any closer
April 05, 2023, 02:41:56 PM
yea I could use iceland2k14 scalar program and make 16k keys and use albertobsd bsgs and load them all takes about 15 minutes to load and get 10 trillion/sec and do random on my 1070 but I am in Florida and the room is already too hot lol. my little 1030 is doing dp 10
April 05, 2023, 02:35:44 PM
I can do 1 quintillion/sec 1,000,000,000,000,000,000 but the range of 110 alone is 649037107316853453566312041152512 or
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
I have. I have over 33GB of tame points (DP 25) in the 110 bit range lol. But I wouldn’t chase down a 4 character h160 that’s for sure lol. six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
Also, the “1 hour” entire range scan had me laughing 😂
April 05, 2023, 02:20:38 PM
I can do 1 quintillion/sec 1,000,000,000,000,000,000 but the range of 110 alone is 649037107316853453566312041152512 or
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
six hundred forty nine nonillion ,
thirty seven octillion ,
one hundred seven septillion ,
three hundred sixteen sextillion ,
eight hundred fifty three quintillion ,
four hundred fifty three quadrillion ,
five hundred sixty six trillion ,
three hundred twelve billion ,
forty one million ,
one hundred fifty two thousand ,
five hundred twelve
so it would take me 649,037,107,316,853 seconds or 7,635,730,674 days 20,919,810 (a little over 20 million years, I better stop smoking) just to go all the way through 110 with my 1070 and a whole lot of luck. I will try with kangaroo with an old gt1030 which does about 370 Trillion/ sec and about as much electricity as an led bulb. have you scanned any of 110?
April 05, 2023, 02:17:06 PM
hi guys, im working on puzzle 125 for a long time, today, i have substracted the public key (im always test and substract by random) of 125 to a smaller range randomly i found a substracted key his hash160 start with the same 4 lettres as a hash160 of an adress in the range 110 bit, so, am i close to find it ? if i search this substracted public key in 110 bit range will i found it ? the probleme im runing 7TerraK per sec its slow, and if anyone can help we will split the reward of the puzzle 125,
02a0397a2b74177e0f101ab1cbe9d560ac384c01fd69d7e19593d1ab3155f9ad73
2000000000000000000000000000:3fffffffffffffffffffffffffff
if you have better speed search from 1f00000000000000000000000000:7fffffffffffffffffffffffffff from 109 to 111 bits,
if you have good speed and can finish this range in 1hour, you wont lose nothing, we will split the reward thank you
How can you search for a 125 bit range public key in 109-111 bit range? And why would you care about the first 4 characters of a hash160 of another random address?02a0397a2b74177e0f101ab1cbe9d560ac384c01fd69d7e19593d1ab3155f9ad73
2000000000000000000000000000:3fffffffffffffffffffffffffff
if you have better speed search from 1f00000000000000000000000000:7fffffffffffffffffffffffffff from 109 to 111 bits,
if you have good speed and can finish this range in 1hour, you wont lose nothing, we will split the reward
thank you Please explain the magic here.😉
If you shrink #125 down to 110 bits, you will create 2^15 new public keys. One of those will be guaranteed to be in the 110 bit range.
I do not understand the initial post of finding the first 4 hash160 characters...IMO, that does not help nor narrow down the search to that one pubkey.
April 05, 2023, 01:52:48 PM
hi guys, im working on puzzle 125 for a long time, today, i have substracted the public key (im always test and substract by random) of 125 to a smaller range randomly i found a substracted key his hash160 start with the same 4 lettres as a hash160 of an adress in the range 110 bit, so, am i close to find it ? if i search this substracted public key in 110 bit range will i found it ? the probleme im runing 7TerraK per sec its slow, and if anyone can help we will split the reward of the puzzle 125,
02a0397a2b74177e0f101ab1cbe9d560ac384c01fd69d7e19593d1ab3155f9ad73
2000000000000000000000000000:3fffffffffffffffffffffffffff
if you have better speed search from 1f00000000000000000000000000:7fffffffffffffffffffffffffff from 109 to 111 bits,
if you have good speed and can finish this range in 1hour, you wont lose nothing, we will split the reward thank you
How can you search for a 125 bit range public key in 109-111 bit range? And why would you care about the first 4 characters of a hash160 of another random address?02a0397a2b74177e0f101ab1cbe9d560ac384c01fd69d7e19593d1ab3155f9ad73
2000000000000000000000000000:3fffffffffffffffffffffffffff
if you have better speed search from 1f00000000000000000000000000:7fffffffffffffffffffffffffff from 109 to 111 bits,
if you have good speed and can finish this range in 1hour, you wont lose nothing, we will split the reward
thank you Please explain the magic here.😉
April 05, 2023, 01:08:16 PM
hi guys, im working on puzzle 125 for a long time, today, i have substracted the public key (im always test and substract by random) of 125 to a smaller range randomly i found a substracted key his hash160 start with the same 4 lettres as a hash160 of an adress in the range 110 bit, so, am i close to find it ? if i search this substracted public key in 110 bit range will i found it ? the probleme im runing 7TerraK per sec its slow, and if anyone can help we will split the reward of the puzzle 125,
02a0397a2b74177e0f101ab1cbe9d560ac384c01fd69d7e19593d1ab3155f9ad73
2000000000000000000000000000:3fffffffffffffffffffffffffff
if you have better speed search from 1f00000000000000000000000000:7fffffffffffffffffffffffffff from 109 to 111 bits,
if you have good speed and can finish this range in 1hour, you wont lose nothing, we will split the reward thank you
02a0397a2b74177e0f101ab1cbe9d560ac384c01fd69d7e19593d1ab3155f9ad73
2000000000000000000000000000:3fffffffffffffffffffffffffff
if you have better speed search from 1f00000000000000000000000000:7fffffffffffffffffffffffffff from 109 to 111 bits,
if you have good speed and can finish this range in 1hour, you wont lose nothing, we will split the reward
thank you March 08, 2023, 12:29:00 AM
Guys I'm telling you, Satoshi solved the puzzle, few days after I pointed him to this thread, he did it with his super GPU computer, and I bet he is busy with the next one #125. I just didn't expect him to take the loots alone, we were partners.🤣 at least let me loot the garbage coins. Lol.
March 07, 2023, 10:02:21 AM
Not a 32 times bigger but √32 like ~5.65. this not full space search, this birthday collision check.
You can compare 'expected operations' for
115 2^58.1
120 2^60.61
125 2^63.10
130 2^65.61
...
You can compare 'expected operations' for
115 2^58.1
120 2^60.61
125 2^63.10
130 2^65.61
...
March 07, 2023, 08:43:27 AM
Hi @zielar, nice to see you here
As we know the puzzle 125 is 32 times bigger than puzzle 120 but prize is only a 4% bigger, i doubt about that such "magic" tool exists. But we don't have way to know.
If that exists it should be a mathematical approach.
2. Someone has a tool to find #125 soon
As we know the puzzle 125 is 32 times bigger than puzzle 120 but prize is only a 4% bigger, i doubt about that such "magic" tool exists. But we don't have way to know.
If that exists it should be a mathematical approach.
Nice to see you too :-)
Despite the passage of time and the fact that people who are more active in the subject know each other only from here - there is a smile on your face knowing that everything is fine with you :-)
It is also certain that no matter how much time passes - the solution of the next thresholds will certainly take place, whether it is due to the fact that the technology will develop further or because the value of assets, despite being only 4% more - will be much higher in terms of physical currency than today. Time will tell. I also believe that we will get to know the finder of solution 120 and how he did it, how much time he needed for it, or at least the private key to complete the list of keys of solved addresses. Greetings!
March 07, 2023, 08:26:53 AM
Hi @zielar, nice to see you here
As we know the puzzle 125 is 32 times bigger than puzzle 120 but prize is only a 4% bigger, i doubt about that such "magic" tool exists. But we don't have way to know.
If that exists it should be a mathematical approach.
We already know that puzzles a random and there is not math relationship between them
2. Someone has a tool to find #125 soon
As we know the puzzle 125 is 32 times bigger than puzzle 120 but prize is only a 4% bigger, i doubt about that such "magic" tool exists. But we don't have way to know.
If that exists it should be a mathematical approach.
I made some computations. Used Python 3. I was curious if the next puzzle can be ax+b=y.
d1, d2, d3 are corresponding private keys from puzzle. Needed to compute a and b.
d1, d2, d3 are corresponding private keys from puzzle. Needed to compute a and b.
We already know that puzzles a random and there is not math relationship between them
March 06, 2023, 01:05:56 PM
I made some computations. Used Python 3. I was curious if the next puzzle can be ax+b=y.
d1, d2, d3 are corresponding private keys from puzzle. Needed to compute a and b.
First script for puzzle 66:
Second script for puzzle 120:
The output is however strange, because a=b!
This means the next puzzle is multiplied and added to it the same value, I mean the multiplicator and later adding a number are the same.
First script gives:
Second script gives:
What can mean that just raising the value(s) of a and b can give next puzzle private key!
d1, d2, d3 are corresponding private keys from puzzle. Needed to compute a and b.
First script for puzzle 66:
Code:
#!/usr/bin/env python3
import math
d1=0x0000000000000000000000000000000000000000000000007cce5efdaccf6808
d2=0x000000000000000000000000000000000000000000000000f7051f27b09112d4
d3=0x000000000000000000000000000000000000000000000001a838b13505b26867
asqr=(d2-(d2/d1))/d1
a=math.sqrt(asqr)
print("a =",a)
b=d2/(a*d1)
print("b =",b)
d4=a*d3+b
print("next =",hex(int(d4)))
import math
d1=0x0000000000000000000000000000000000000000000000007cce5efdaccf6808
d2=0x000000000000000000000000000000000000000000000000f7051f27b09112d4
d3=0x000000000000000000000000000000000000000000000001a838b13505b26867
asqr=(d2-(d2/d1))/d1
a=math.sqrt(asqr)
print("a =",a)
b=d2/(a*d1)
print("b =",b)
d4=a*d3+b
print("next =",hex(int(d4)))
Second script for puzzle 120:
Code:
#!/usr/bin/env python3
import math
d1=0x000000000000000000000000000000000000016f14fc2054cd87ee6396b33df3
d2=0x00000000000000000000000000000000000035c0d7234df7deb0f20cf7062444
d3=0x0000000000000000000000000000000000060f4d11574f5deee49961d9609ac6
asqr=(d2-(d2/d1))/d1
a=math.sqrt(asqr)
print("a =",a)
b=d2/(a*d1)
print("b =",b)
d4=a*d3+b
print("next =",hex(int(d4)))
import math
d1=0x000000000000000000000000000000000000016f14fc2054cd87ee6396b33df3
d2=0x00000000000000000000000000000000000035c0d7234df7deb0f20cf7062444
d3=0x0000000000000000000000000000000000060f4d11574f5deee49961d9609ac6
asqr=(d2-(d2/d1))/d1
a=math.sqrt(asqr)
print("a =",a)
b=d2/(a*d1)
print("b =",b)
d4=a*d3+b
print("next =",hex(int(d4)))
The output is however strange, because a=b!
This means the next puzzle is multiplied and added to it the same value, I mean the multiplicator and later adding a number are the same.
First script gives:
Code:
a = 1.4068509690560507
b = 1.4068509690560507
next = 0x254d0fd348f800000
b = 1.4068509690560507
next = 0x254d0fd348f800000
Second script gives:
Code:
a = 6.122671382779787
b = 6.122671382779785
next = 0x251a1b44dd0e160000000000000000
b = 6.122671382779785
next = 0x251a1b44dd0e160000000000000000
What can mean that just raising the value(s) of a and b can give next puzzle private key!
March 05, 2023, 09:01:05 PM
It is a real shame that the solver of #120 still didn't give us the private key.
Dude, if you're out there, please find a way to communicate it
Dude, if you're out there, please find a way to communicate it
There are various reasons why we still don't have the key:
1. The finder wishes to remain anonymous
2. Someone has a tool to find #125 soon
3. The withdrawal was made by someone outside the forum
4. Someone has some other purpose in it, so as not to reveal themselves.
It is certainly puzzling that BCH and BTG have not been paid out from this wallet so far, while the payment was made to the address without any other transaction before, nor was it sent further - which also raises the suspicion that the finder could use some tool that worked for a long time -> found the right key and made the withdrawal, but the initiator himself still does not know that he found solution #120. A lot of unknowns, but I believe that eventually someone will somehow provide a solution, because the solution #120 without indicating the key to the public will simply be a blow to the whole history of these puzzles.
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